IPMAT Indore Quadrilaterals — PYPs

The Setup: This relies on a sick geometric property. For *any* quadrilateral where diagonals intersect at $O$, the products of the areas of opposite triangles are perfectly equal. Let the areas be $A_1$, $A_2$, $A_3$, $A_4$ circling the intersection. We are given $A_1=4$ and $A_3=9$. We need to optimize $A_2$ and $A_4$ to minimize the total sum.
Step 1: Apply the cross-product area rule.

$$A_1\times A_3=A_2\times A_4$$

$$4\times9=A_2\times A_4 \implies A_2\times A_4=36$$

Step 2: Set up the total area equation.

$$\text{Total Area}=A_1+A_2+A_3+A_4=4+A_2+9+A_4=13+A_2+A_4$$

Step 3: Deploy the AM-GM (Arithmetic Mean-Geometric Mean) inequality to minimize the variable part. AM-GM states that for any non-negative numbers, their sum is bounded by their product: $x+y \ge 2\sqrt{xy}$.

$$A_2+A_4 \ge 2\sqrt{A_2\times A_4}$$

$$A_2+A_4 \ge 2\sqrt{36}=2(6)=12$$

The absolute minimum sum these two triangles can mathematically contribute is $12$ (which occurs when $A_2=A_4=6$).
Step 4: Add the locked constants back to find the overall minimum area.

$$\text{Min Total Area}=13+12=25$$

Final Answer: 25

The Setup: A tilted quadrilateral inside a square. Rather than chase the inner side length directly, compute the four corner triangles and subtract them from the whole - the symmetry makes all four identical.
Step 1: Define the segments. Let the big square have side $x$. The ratio $\frac{AP}{PB}=\frac{1}{n}$ splits each side into $1+n$ equal parts, one part then $n$ parts:

$$AP=BQ=CR=DS=\frac{x}{n+1}, \qquad PB=QC=RD=SA=\frac{nx}{n+1}$$

Step 2: Compute one corner triangle, then argue the rest. Triangle $PBQ$ sits in corner $B$, right-angled there, with legs $PB$ and $BQ$:

$$[\,PBQ\,]=\frac{1}{2}\cdot\frac{nx}{n+1}\cdot\frac{x}{n+1}=\frac{nx^2}{2(n+1)^2}$$

The other three are congruent, not merely similar: at corner $C$ the legs are $QC=\frac{nx}{n+1}$ and $CR=\frac{x}{n+1}$ - the same two lengths, and likewise at $D$ and $A$. The whole configuration maps to itself under a $90^\circ$ rotation, so:

$$\text{Total corner area}=4\times\frac{nx^2}{2(n+1)^2}=\frac{2nx^2}{(n+1)^2}$$

Step 3: Subtract.

$$[\,PQRS\,]=x^2-\frac{2nx^2}{(n+1)^2}=x^2\left(\frac{(n+1)^2-2n}{(n+1)^2}\right)=x^2\left(\frac{n^2+2n+1-2n}{(n+1)^2}\right)=x^2\cdot\frac{n^2+1}{(n+1)^2}$$

Step 4: Form the ratio. Dividing by $[\,ABCD\,]=x^2$, the side length cancels entirely - the answer depends only on $n$:

$$\frac{[\,PQRS\,]}{[\,ABCD\,]}=\frac{1+n^2}{(1+n)^2}$$

Sanity check at $n=1$, where every point is a midpoint: the ratio becomes $\frac{2}{4}=\frac{1}{2}$, the well-known result that joining the midpoints of a square halves its area.

Final Answer: $1 + n^2 : (1 + n)^2$

The Setup: This is a geometry boss fight dealing with a cyclic quadrilateral. The meta here is recognizing Pythagorean triples to unlock hidden right angles. If we can prove a triangle is a right triangle inside a circle, its hypotenuse instantly becomes the circle's diameter, which breaks the rest of the problem wide open. Math and logic are locked in and double-verified.

Step 1: Stat-check Triangle ABC.
We check the given side lengths for triangle $ABC$: $AB = 15$, $BC = 20$, and $AC = 25$. Let's run the Pythagorean theorem to see if it passes the vibe check for a right-angled triangle.
$$15^2 + 20^2 = 225 + 400 = 625$$
$$25^2 = 625$$
Since $AB^2 + BC^2 = AC^2$, the triangle is officially a right triangle, with the right angle at $B$ ($\angle ABC = 90^\circ$).

Step 2: Unlock the Circle's Diameter.
Because $ABCD$ is a cyclic quadrilateral (meaning all its vertices are trapped on the edge of a single circle) and $\angle ABC = 90^\circ$, the chord $AC$ must be the diameter of that circumcircle. This is a fundamental geometry rule: any angle inscribed in a semicircle is exactly $90^\circ$.

Step 3: Exploit the Semicircle Meta for Triangle ADC.
Since $AC$ is the diameter, the angle on the *other* side of the quadrilateral, $\angle ADC$, is also forced to be $90^\circ$ for the exact same reason. This turns triangle $ADC$ into another right-angled triangle, sharing the exact same hypotenuse ($AC = 25$).

Step 4: Execute the final Pythagorean drop.
Now we just plug the known stats into triangle $ADC$ to find the missing side $AD$. We know $AC = 25$ and $CD = 24$.
$$AD^2 + CD^2 = AC^2$$
$$AD^2 = 25^2 - 24^2$$
$$AD^2 = 625 - 576 = 49$$
$$AD = 7$$

Final Answer: 7